Let the side of right angled triangles are x,x+1,x+2 ∴(x+2)2=(x+1)2+(x)2 ⇒x2+4x+4=x2+2x+1+x2 ⇒x2−2x−3=0 ⇒(x−3)(x+1)=0 ⇒x=3,x≠−1 Perimeter of triangle =3x+3 3x+3<30 x+1<10 x<9 The possible triangle whose sides are (3,4,5),(6,8,10) ∴ Two triangles are possible.